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%I #29 Mar 14 2024 11:49:50
%S 1,14,391,10934,305761,8550374,239104711,6686381534,186979578241,
%T 5228741809214,146217791079751,4088869408423814,114342125644787041,
%U 3197490648645613334,89415396036432386311,2500433598371461203374,69922725358364481308161,1955335876435834015425134
%N Chebyshev T-polynomials T(n,14) with Diophantine property.
%C a(n)^2 - 195 b(n)^2 = +1 with b(n):=A097311(n) gives all nonnegative solutions of this Pell equation.
%C a(195+390k)-1 and a(195+390k)+1 are consecutive odd powerful numbers. See A076445. - _T. D. Noe_, May 04 2006
%C Except for the first term, positive values of x (or y) satisfying x^2 - 28xy + y^2 + 195 = 0. - _Colin Barker_, Feb 23 2014
%H Vincenzo Librandi, <a href="/A097310/b097310.txt">Table of n, a(n) for n = 0..700</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (28,-1).
%F a(n) = 28*a(n-1) - a(n-2), a(-1):= 14, a(0)=1.
%F a(n) = T(n, 14)= (S(n, 28)-S(n-2, 28))/2 = S(n, 28)-14*S(n-1, 28) with T(n, x), resp. S(n, x), Chebyshev's polynomials of the first, resp.second, kind. See A053120 and A049310. S(n, 28)=A097311(n).
%F a(n) = (ap^n + am^n)/2 with ap := 14+sqrt(195) and am := 14-sqrt(195).
%F a(n) = sum(((-1)^k)*(n/(2*(n-k)))*binomial(n-k, k)*(2*14)^(n-2*k), k=0..floor(n/2)), n>=1.
%F G.f.: (1-14*x)/(1-28*x+x^2).
%F a(n) = sqrt(1 + 195*A097311(n)^2), n>=0.
%t LinearRecurrence[{28,-1},{1,14},20] (* _Harvey P. Dale_, Jan 29 2014 *)
%t CoefficientList[Series[(1 - 14 x)/(1 - 28 x + x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Feb 24 2014 *)
%o (Sage) [lucas_number2(n,28,1)/2 for n in range(0,16)] - _Zerinvary Lajos_, Jun 27 2008
%o (PARI) Vec((1-14*x)/(1-28*x+x^2) + O(x^100)) \\ _Colin Barker_, Feb 23 2014
%Y Cf. A090249, A097311.
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Aug 31 2004
%E More terms from _Colin Barker_, Feb 23 2014
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